Question

if initially 68g ammonia and 192g oxygen is present , calculate the app. mass % of NO(g) formed at 80% completion of the reaction

Answer 1

4NH3 + 5O2 ------> 4NO + 6H2O

NH3 = 14 + 3 = 17

Moles of NH3 = 68/17 = 4

Moles of O2 = 192/32 = 6

From the equation, 4 moles of NH3 react with 5 moles of O2, meaning O2 is in excess. NH3 is the limiting reagent

From the equation, 4 moles NH3 will react to produce 4 moles of NO

Molar mass of NO = 14 + 16 = 20

Mass of NO produced at 80% completion of the reaction = 80% x 4 x 20

                                                                                            = 64g

Answer 2

Explanation:

Ag3A = 3Ag + A

3 moles of Ag combine with 3 moles of A

Mass of Ag = 0.37

Mass of A = 607 - 0.37 = 606.63

Moles of Ag = 0.37/108 = 0.003426

Moles of A = 0.003426/3 = 0.001142

Molar mass of A = 606.63/0.001142 = 531211

A tribasic acid has 3 atom of H

Molar mass of the acid = 3 + 531211 = 531214 g/mol